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Question: One vertex of an equilateral triangle is \( (2,3) \) and the equation of the line opposite to the ve...

One vertex of an equilateral triangle is (2,3)(2,3) and the equation of the line opposite to the vertex is x+y=2,x + y = 2, equations of the remaining two sides are:
A. y−3=±2(x+2)y - 3 = \pm 2(x + 2)
B. y−3=(3±1)(x−2)y - 3 = (\sqrt 3 \pm 1)(x - 2)
C. y−3=(2±3)(x−2)y - 3 = (2 \pm \sqrt 3 )(x - 2)
D.None

Explanation

Solution

Hint : Here, use the measure of angle equal to 60∘60^\circ since the given triangle is equilateral. Also, use the standard formula for the line equation and the standard quadratic equation and find the roots of the equation and find the line passing through the points.

Complete step-by-step answer :
Given,
x+y=2x + y = 2
Convert the given equation in the standard form –
y=mx+cy = mx + c
⇒y=−x+2\Rightarrow y = - x + 2
The slope, m=(−1)m = ( - 1)
In an equilateral triangle all the angles are equal with measures equal 60∘60^\circ .
If m is the slope of a line which makes an angle 60∘60^\circ with the above line, then
tan⁡60∘=∣m1−m21+m1m2∣\tan 60^\circ = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|
Where,
m1=m m2=(−1)   {m_1} = m \\\ {m_2} = ( - 1) \;
Also, tan⁡60∘=3\tan 60^\circ = \sqrt 3
Place the values in the above equation –
⇒3=∣m+11−m∣\Rightarrow \sqrt 3 = \left| {\dfrac{{m + 1}}{{1 - m}}} \right|
Apply squares on both the sides of the equation –
⇒(3)2=(∣m+11−m∣)2\Rightarrow {\left( {\sqrt 3 } \right)^2} = {\left( {\left| {\dfrac{{m + 1}}{{1 - m}}} \right|} \right)^2}
Simplify the above equation-
⇒3=(m+11−m)2\Rightarrow 3 = {\left( {\dfrac{{m + 1}}{{1 - m}}} \right)^2}
Powers on the right hand the side of the equation can be given on the numerators and the denominators separately.
⇒3=(m+1)2(1−m)2\Rightarrow 3 = \dfrac{{{{\left( {m + 1} \right)}^2}}}{{{{\left( {1 - m} \right)}^2}}}
Do cross- multiplication. Where, the denominator is multiplied with the numerator on the opposite side.
⇒3(1−m)2=(m+1)2\Rightarrow 3{\left( {1 - m} \right)^2} = {\left( {m + 1} \right)^2}
Apply the whole square formula in the above equation-
⇒3(1−2m+m2)=(m2+2m+1)\Rightarrow 3\left( {1 - 2m + {m^2}} \right) = \left( {{m^2} + 2m + 1} \right)
Open the brackets and apply multiply the number outside the bracket.
⇒(3−6m+m2)=(m2+2m+1)\Rightarrow \left( {3 - 6m + {m^2}} \right) = \left( {{m^2} + 2m + 1} \right)
Take all the terms on one side of the equation. Remember when the term is moved from one side to another, the sign of the term is also changed. Positive terms become negative and vice-versa.

⇒3−6m+3m2=m2+2m+1 ⇒3−6m+3m2−m2−2m−1=0   \Rightarrow 3 - 6m + 3{m^2} = {m^2} + 2m + 1 \\\ \Rightarrow 3 - 6m + 3{m^2} - {m^2} - 2m - 1 = 0 \;

Make pairs of the like terms –
⇒3m2−m2‾−2m−6m‾−1+3‾=0\Rightarrow \underline {3{m^2} - {m^2}} - \underline {2m - 6m} - \underline {1 + 3} = 0
Add/Subtract between the like terms –
⇒2m2−8m+2=0\Rightarrow 2{m^2} - 8m + 2 = 0
Compare the above equation –
ax2+bx+c=0a{x^2} + bx + c = 0 and x=−b+Δ2ax = \dfrac{{ - b + \sqrt \Delta }}{{2a}} where Δ=b2−4ac\Delta = {b^2} - 4ac
⇒m=2±3\Rightarrow m = 2 \pm \sqrt 3
Now, the equation of the two sides passing through (2,3)(2,3) are-
⇒y−3=(2+3)(x−2) and y−3=(2−3)(x−2)   \Rightarrow y - 3 = \left( {2 + \sqrt 3 } \right)(x - 2){\text{ and}} \\\ y - 3 = \left( {2 - \sqrt 3 } \right)(x - 2) \;
So, the correct answer is “Option C”.

Note : Be careful while simplification of the equation and while finding the roots of the equation. Remember to give appropriate signs when the terms are moved from one side to another. Remember the trigonometric table for the values for the easy substitution and an accurate answer.